Uniform integrability of the stopped compensated Poisson process

Question

Let $N(t)$ be a Poisson process of rate $\lambda$ and consider the compensated Poisson process

$$\bar{N}(t):= N(t) - \lambda t.$$

It was already shown in another post (Is a compensated Poisson process uniformly integrable) that $\bar{N}(t)$ is not uniformly integrable.

Consider a stopping time $T$ and assume that $\mathbb E T < \infty$. Is it always true that the stopped compensated Poisson process process $\bar{N}(t \wedge T)$ is uniformly integrable? If not, what further assumptions on $T$ would imply such result?

Answer 1

Yes, the stopped compensated Poisson process is uniformly integrable:

By Doob's maximal inequality, we have

$$\mathbb{E} \left( \sup_{t \in [0,K]} |\bar{N}(t \wedge T)|^2 \right) \leq 4 \mathbb{E}(|\bar{N}(K \wedge T)|^2)$$

for any $K>0$. Since $(\bar{N}_t^2-\lambda t)_{t \geq 0}$ is a martingale, we obtain

$$\mathbb{E} \left( \sup_{t \in [0,K]} |\bar{N}(t \wedge T)|^2 \right) \leq 4\lambda \mathbb{E}(K \wedge T) \leq 4 \lambda \mathbb{E}(T).$$

Letting $K \to \infty$ (using monotone convergence), we get

$$\mathbb{E} \left( \sup_{t \geq 0}|\bar{N}(t \wedge T)|^2 \right)<\infty$$ and this implies uniform integrability.